Question

The spring of the pressure gauge shown in Figure P14.7 has a force constant of $1250N/m$ , and the piston has a diameter of $1.20\text{cm}$ . As the gauge is lowered into the water in a lake, what change in depth causes the piston to move in by$0.750\text{cm}$ ?

Short answer

The change in depth causes the piston to move in by 0.750 cm is$h=8.46m$ .

Step-by-step solution

A concept:

Pascal’s law states that when pressure is applied to an enclosed fluid, the pressure is transmitted undiminished to every point in the fluid and to every point on the walls of the container.

The pressure in a fluid at rest varies with depth h in the fluid according to the expression

$P=\left(P_0+\rho gh\right)$

Where, $P_0$is the pressure at role="math" localid="1663666065886" $h=0$and p is the density of the fluid, assumed uniform.

Hooke’s law states that the applied force equals a constant times the displacement or change in length .

$F=Kx$

The value of k depends not only on the kind of elastic material under consideration but also on its dimensions and shape.

Determine the change in depth: 

Assuming the spring obeys Hooke’s law, the increase in force on the piston required to compress the spring an additional amount $\Delta x$ is,

$$\begin{gathered} \Delta F=F-F_0 \\ =\left(P-P_0\right)A \\ =k\left(\Delta x\right) \end{gathered}$$

The gauge pressure at a depth of h beneath the surface of a fluid is,

$\left(P-P_0\right)=\rho gh$

So, you have,

$$\begin{gathered} \rho ghA=k\left(\Delta x\right) \\ h=\frac{k\left(\Delta x\right)}{\rho gA} \end{gathered}$$

If$k=1250N/m,A=\frac{\pi d^2}{4},d=1.20\times {10}^{-2}m$the fluid is water so the density is $\left(\rho =1.00\times {10}^{3}kg/m^3\right)$and the depth required to compress the spring an additional is $\Delta x=0.750\times {10}^{-2}m$. Therefore,

$$\begin{gathered} h=\frac{{\displaystyle \left(1250N/m\right)\left(0.750m\right)}}{\left(1.00\times {10}^{3}Kg/m^3\right){\displaystyle \left(9.8m/s^2\right)}{\displaystyle \left({\frac{{\displaystyle 3.14\times \left(1.20\times {10}^{-2}m\right)}}{4}}^2\right)}} \\ =8.46m \end{gathered}$$